Problem: If $AAA_4$ can be expressed as $33_b$, where $A$ is a digit in base 4 and $b$ is a base greater than 5, what is the smallest possible sum $A+b$?
Answer: We can rewrite $AAA_4$ and $33_b$ to get \begin{align*}
16A+4A+A&=3b+3\quad\Rightarrow\\
21A&=3b+3.
\end{align*}The smallest possible value for $A$ is 1, which gives us $21=3b+3$ and $b=6$. So the smallest sum $A+b=\boxed{7}$. While there are other values for $A$ and $b$ that work, increasing $A$ will increase $b$, resulting in a larger sum.